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Question

The ionization constant of dimethylamine is 5.4×104. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

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Solution

Kb=5.4×104c=0.02M
Then α=Kbc=5.4×1040.02=0.1643
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

NaOH(aq)Na+(aq)+OHaq0.1M0.1M
And,
(CH3)2NH+H2O(CH3)2NH+2+OH(0.02x)xx0.02M0.1M
Then, [(CH3)2NH+2]=X[OH]=x+0.1Kb=[(CH3)2NH+2][OH][(CH3)2NH]5.4×104=x×0.10.02x=0.0054
It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.


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